Why we need a second law
The first law allows many things that never happen. A cup of tea could get hotter by taking heat from the cooler room; energy would still be conserved. A ship could run by taking heat from the sea and turning all of it into work. These never happen.
The second law of thermodynamics tells us the direction in which natural processes go, and sets a limit on how much heat can become work.
Heat engines and efficiency
A heat engine is a device that turns heat into work again and again, working in a cycle. It has three parts:
- a hot reservoir (source) at T₁, which gives heat Q₁;
- a working substance (gas, steam) that goes round a cycle and does work W;
- a cold reservoir (sink) at T₂, which receives the rejected heat Q₂.
In a cycle ΔU = 0, so by the first law W = Q₁ − Q₂.
Efficiency η = W/Q₁ = 1 − Q₂/Q₁.
Examples: steam engine, petrol and diesel engines, thermal power station.
Refrigerators and heat pumps
A refrigerator is a heat engine run backwards. Work W from the compressor makes the working gas take heat Q₂ from the cold inside and give heat Q₁ = Q₂ + W to the warm room.
Coefficient of performance α = Q₂/W = Q₂/(Q₁ − Q₂).
α can be more than 1 (typical fridges: 2–6), but it can never be infinite, because W can never be zero. A heat pump uses the same idea to warm a room with heat taken from cold outside air.
Statements of the second law
Kelvin–Planck statement
No process is possible whose only result is to take heat from a reservoir and turn it completely into work. In other words, no engine can have 100% efficiency; it must reject some heat Q₂ to a colder body.
Clausius statement
No process is possible whose only result is to move heat from a colder body to a hotter body. In other words, a perfect refrigerator (W = 0) is impossible; heat flows uphill only when work is done.
Both say the same thing
If a perfect engine existed, we could use its work to run a fridge and the pair together would move heat from cold to hot with no outside work, breaking Clausius. So breaking one statement breaks the other: they are equivalent.
Carnot engine and Carnot's theorem
The Carnot engine is an ideal reversible engine between a source at T₁ and a sink at T₂. Its working gas goes round the Carnot cycle:
- isothermal expansion at T₁ (takes Q₁),
- adiabatic expansion (T₁ → T₂),
- isothermal compression at T₂ (gives out Q₂),
- adiabatic compression (T₂ → T₁), back to the start.
For this cycle Q₂/Q₁ = T₂/T₁, so
η(Carnot) = 1 − T₂/T₁ (T in kelvin).
Carnot's theorem: (a) no engine between two given temperatures can be more efficient than a Carnot engine; (b) all reversible engines between the same two temperatures have the same efficiency, whatever the working substance.
η = 1 only if T₂ = 0 K, which cannot be reached, so 100% efficiency is impossible. For a Carnot fridge, α = T₂/(T₁ − T₂).
Exam tip: 2–3 mark questions ask for the two statements, efficiency/COP numericals, and the four steps of the Carnot cycle.
Try it
Put your hand near the back grill of a running fridge (do not touch the wires). It is warm: that is Q₁ = Q₂ + W being dumped into your kitchen. In the 3D, set T₂ close to T₁ and watch the Carnot efficiency drop towards zero.
Key formulas and definitions
- Engine: W = Q₁ − Q₂
- Efficiency η = W/Q₁ = 1 − Q₂/Q₁
- Carnot: η = 1 − T₂/T₁ (T in K)
- Carnot: Q₂/Q₁ = T₂/T₁
- Refrigerator: α = Q₂/W = Q₂/(Q₁ − Q₂)
- Carnot refrigerator: α = T₂/(T₁ − T₂)
Worked examples
1. An engine takes 1000 J of heat and does 300 J of work per cycle. Find its efficiency and the heat rejected.
η = W/Q₁ = 300/1000 = 0.3 = 30%. Q₂ = Q₁ − W = 700 J.
2. An engine rejects 600 J for every 800 J it takes. Find its efficiency.
η = 1 − Q₂/Q₁ = 1 − 600/800 = 0.25 = 25%.
3. Find the efficiency of a Carnot engine working between 127 °C and 27 °C.
T₁ = 400 K, T₂ = 300 K. η = 1 − 300/400 = 0.25 = 25%.
4. A refrigerator takes 300 J from the inside using 100 J of work. Find its COP and the heat given to the room.
α = Q₂/W = 300/100 = 3. Q₁ = Q₂ + W = 400 J.
5. A Carnot engine has an efficiency of 40% with a sink at 300 K. Find the source temperature.
0.4 = 1 − 300/T₁ ⇒ 300/T₁ = 0.6 ⇒ T₁ = 500 K (227 °C).
6. A Carnot fridge keeps its freezer at −23 °C in a 27 °C room. Find its COP. How much work removes 5000 J from the freezer?
T₂ = 250 K, T₁ = 300 K. α = T₂/(T₁ − T₂) = 250/50 = 5. W = Q₂/α = 5000/5 = 1000 J.
7. An inventor claims an engine between 500 K and 300 K that takes 1000 J and gives 450 J of work. Is it possible?
Best possible (Carnot) η = 1 − 300/500 = 40%, so at most 400 J of work. The claim of 450 J (45%) breaks the second law. Not possible.
Common mistakes
- Using °C in η = 1 − T₂/T₁. Always convert to kelvin first.
- Thinking a better engine design can reach 100% efficiency. The second law forbids it for any engine.
- Mixing up engine efficiency (W/Q₁, always less than 1) and fridge COP (Q₂/W, can be more than 1).
- Thinking the Clausius statement means heat can never go from cold to hot. It can, but only when work is supplied (fridge, AC).